Questions
Question 1a
Ha: The mean for all the groups are different
H0: µ1 = µ2 = µ3 = µ4
Ha: µ1 ≠ µ2 ≠ µ3 ≠µ4
Question 1b
H02: µ1 = (µ2 +µ3 + µ4)
H03: µ2 = (µ3 + µ4)
H04: µ3 = µ4
Therefore:
H01: (0) µ1 + (0) µ2 + (0) µ3 + (0) µ4
H02: (1) µ1 – (1/3) µ2 - (1/3) µ3 - (1/3) µ4
H03: (0) µ1 + (1) µ2 - (1/2) µ3 - (1/2) µ4
H03: (0) µ1 + (0) µ2 + (1) µ3 - (1) µ4
Question 2
Property 1
H01: (0) + (0) + (0) + (0) = 0
H02: (1) – (1/3) - (1/3) - (1/3) = 0
H03: (0) + (1) - (1/2) - (1/2) = 0
H04: (0) + (0) + (1) - (1) = 0
Property 2
The sum of the product for the coefficients in pairwise comparison = 0
Contrast 2 vs 3
(1)(0) – (1/3) (1) + (1/3) (1/2) + (1/3) (1/2) = 0
Contrast 2 vs 4
(1)(0) – (1/3) (0) - (1/3) (1) + (1/3) (1) = 0
Contrast 3 vs 4
(0)(0) + (1) (0) - (1/2) (1) + (1/2) (1) = 0
The first assumption of orthogonality is that the linear sum is zero which is proven in property 1. The second assumption is The sum of the product for the coefficients in pairwise comparison is zero, which is proven in property 2. Therefore, the contrast are orthogonal.
Question 3
An analysis of variance revealed that the effect of treatment was significant (F 3,36) = 20.2 p < 0.0001. Bonferroni test revealed that there is statistically significant difference between the control group (no treatment) and treatment group (opportunity to cheat). Bonferroni p-value = 0.0010479 < 0.01. There was also a significant difference between the group that had an opportunity to cheat without incentive and the groups that had both opportunity and incentive to cheat. Bonferroni p-value = 0.0029336 < 0.01. Similarly, there was a statistically significant difference between the group that had a small incentive to lie and the group that had a large incentive to lie. Bonferroni p-value = 0.0040928 < 0.01.
Question 4
The Bonferroni correction is used in multiple comparison to correct for the numerous independent variables for simultaneous tests. Adjustments are made on the obtained p-values by the dividing the p-value calculated with the number of variables. It was not used then the null hypothesis may have been rejected falsely or may have failed to it reject it falsely. For instance, if the p-value is 0.13 and the number of comparisons are 10, we will fail to reject the null at 10% significance level (p =. 13 > 0.1). However, with correction (0.13/10 = 0.013), we will reject the null at 10% significance level (p =. 013 < 0.1).
Question 5
A repeated measure design uses the same study participants in all the experimental conditions.
It would not have been appropriate for this design. This is because learning may take place. Therefore, the subjects may become better at predicting the outcomes: thus introducing a confounding variable. The observed results may be because the subjects got better at learning and not the incentives provided or a combination of both. The subjects may also get tired as the experiment progresses. Lastly, the ordering effects are bound to also influence the results. Ordering effect occurs when difference arise merely because of the order in which the experiment treatment was conducted.
However, there are a number of benefits that accrue form a repeated measure design. Firstly, repeated measure design improves eliminates the effects due to between subjects due to physical, genetic, historical factors or other dimension that could create competing hypotheses for the observed effects. Therefore, it is a more powerful statistical tool. Secondly, it allows the experimenter to use fewer subjects because of the enhanced statistical power and the fact that each subject is involved in multiple treatments. For instance, if an experiment requires 10 subjects for each group, then only 10 subjects will be required for the entire experiment.
Lastly, it is cheaper since fewer subjects are used for the experiment. Therefore, recruitment, training and compensation of subjects is less costly than an independent group design.
